मराठी

Solve the Following Equation For X: `Tan^-1 1/4+2tan^-1 1/5+Tan^-1 1/6+Tan^-1 1/X=Pi/4` - Mathematics

Advertisements
Advertisements

प्रश्न

Solve the following equation for x:

`tan^-1  1/4+2tan^-1  1/5+tan^-1  1/6+tan^-1  1/x=pi/4`

Advertisements

उत्तर

We know

`tan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))`

`thereforetan^-1  1/4+2tan^-1  1/5+tan^-1  1/6+tan^-1  1/x=pi/4`

`=>tan^-1  1/4+tan^-1  1/5+tan^-1  1/5+tan^-1  1/6+tan^-1  1/x=pi/4`

`=>tan^-1((1/4+1/5)/(1-1/4xx1/5))+tan^-1((1/5+1/6)/(1-1/5xx1/6))+tan^-1  1/x=pi/4`

`=>tan^-1((9/20)/(19/20))+tan^-1((11/30)/(29/30))+tan^-1  1/x=pi/4`

`=>tan^-1(9/19)+tan^-1(11/29)+tan^-1  1/x=pi/4`

`=>tan^-1((9/19+11/29)/(1-11/29xx1/x))+tan^-1  1/x=pi/4`

`=>tan^-1 (235/226)+tan^-1  1/x=pi/4`

`=>tan^-1((235/226+1/x)/(1-235/226xx1/x))=pi/4`

`=>(235x+226)/(226x-235)=tan  pi/4`

`=>(235x+226)/(226x-235)=1`

`=>235x+226=226x-235`

`=>9x=-461`

`=>x=-461/9`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Inverse Trigonometric Functions - Exercise 4.14 [पृष्ठ ११६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.14 | Q 8.1 | पृष्ठ ११६

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Write the value of `tan(2tan^(-1)(1/5))`


Find the value of the following: `tan(1/2)[sin^(-1)((2x)/(1+x^2))+cos^(-1)((1-y^2)/(1+y^2))],|x| <1,y>0 and xy <1`


Solve the following for x:

`sin^(-1)(1-x)-2sin^-1 x=pi/2`


​Find the principal values of the following:
`cos^-1(-sqrt3/2)`


`sin^-1(sin  (5pi)/6)`


`sin^-1(sin3)`


`sin^-1(sin4)`


Evaluate the following:

`cos^-1{cos  ((4pi)/3)}`


Evaluate the following:

`cos^-1{cos  (13pi)/6}`


Evaluate the following:

`sec^-1(sec  (5pi)/4)`


Evaluate:

`cos{sin^-1(-7/25)}`


Evaluate:

`cosec{cot^-1(-12/5)}`


Evaluate: 

`cot(sin^-1  3/4+sec^-1  4/3)`


If `sin^-1x+sin^-1y=pi/3`  and  `cos^-1x-cos^-1y=pi/6`,  find the values of x and y.


`sin^-1x=pi/6+cos^-1x`


Solve the following equation for x:

tan−1(x + 2) + tan−1(x − 2) = tan−1 `(8/79)`, x > 0


Solve the following equation for x:

`tan^-1(2+x)+tan^-1(2-x)=tan^-1  2/3, where  x< -sqrt3 or, x>sqrt3`


Sum the following series:

`tan^-1  1/3+tan^-1  2/9+tan^-1  4/33+...+tan^-1  (2^(n-1))/(1+2^(2n-1))`


`sin^-1  5/13+cos^-1  3/5=tan^-1  63/16`


Solve the following:

`cos^-1x+sin^-1  x/2=π/6`


If `sin^-1  (2a)/(1+a^2)+sin^-1  (2b)/(1+b^2)=2tan^-1x,` Prove that  `x=(a+b)/(1-ab).`


What is the value of cos−1 `(cos  (2x)/3)+sin^-1(sin  (2x)/3)?`


Write the value of sin1 (sin 1550°).


What is the principal value of `sin^-1(-sqrt3/2)?`


Write the principal value of \[\cos^{- 1} \left( \cos\frac{2\pi}{3} \right) + \sin^{- 1} \left( \sin\frac{2\pi}{3} \right)\]


Write the value of \[\sec^{- 1} \left( \frac{1}{2} \right)\]


Write the principal value of \[\sin^{- 1} \left\{ \cos\left( \sin^{- 1} \frac{1}{2} \right) \right\}\]


Write the value of  \[\tan^{- 1} \left( \frac{1}{x} \right)\]  for x < 0 in terms of `cot^-1x`


Write the value of  `cot^-1(-x)`  for all `x in R` in terms of `cot^-1(x)`


If \[\cos\left( \tan^{- 1} x + \cot^{- 1} \sqrt{3} \right) = 0\] , find the value of x.

 

Find the value of \[\tan^{- 1} \left( \tan\frac{9\pi}{8} \right)\]


If sin−1 − cos−1 x = `pi/6` , then x = 


It \[\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}\]   (−7), then the value of x is

 


The value of sin \[\left( \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} \right)\] is

 


Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .


Find the value of `sin^-1(cos((33π)/5))`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×